題干：

Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the?lexicographical?order.

After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted in?lexicographical?order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes?lexicographical!

She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the?lexicographical?order. If so, you should find out any such order.

Lexicographical?order is defined in following way. When we compare?s?and?t, first we find the leftmost position with differing characters:?si?≠?ti. If there is no such position (i. e.?s?is a prefix of?t?or vice versa) the shortest string is less. Otherwise, we compare characters?si?and?ti?according to their order in alphabet.

Input

The first line contains an integer?n?(1?≤?n?≤?100): number of names.

Each of the following?n?lines contain one string?namei?(1?≤?|namei|?≤?100), the?i-th name. Each name contains only lowercase Latin letters. All names are different.

Output

If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on).

Otherwise output a single word "Impossible" (without quotes).

Examples

Input

3 rivest shamir adleman

Output

bcdefghijklmnopqrsatuvwxyz

Input

10 tourist petr wjmzbmr yeputons vepifanov scottwu oooooooooooooooo subscriber rowdark tankengineer

Output

Impossible

Input

10 petr egor endagorion feferivan ilovetanyaromanova kostka dmitriyh maratsnowbear bredorjaguarturnik cgyforever

Output

aghjlnopefikdmbcqrstuvwxyz

Input

7 car care careful carefully becarefuldontforgetsomething otherwiseyouwillbehacked goodluck

Output

acbdefhijklmnogpqrstuvwxyz

解題報(bào)告：

? ? 好難啊這題，，不知道為什么都覺得簡(jiǎn)單。。明明很難完全卡上拓?fù)渑判虻臈l件啊兩個(gè)判斷結(jié)束的條件，top不足26 ?或者兩串每一個(gè)字符都分別相同并且后串比前串短。

?

AC代碼：

#include<bits/stdc++.h>using namespace std; char str[105][105]; int n; int head[200],in[200]; int maze[200][200]; char ans[200]; int len[200]; bool vis[50][50]; int cnt,top; struct Edge {int to,w,ne; } e[400]; void add(int u,int v,int w) {e[cnt].to = v;e[cnt].ne = head[u];e[cnt].w = w;head[u] = cnt++; } bool topu() {//預(yù)處理 priority_queue< int , vector<int>, greater<int> > pq;//從小到大排 for(int i = 0; i<26; i++) {if(in[i] == 0 /*&& vis[i]*/) { // printf("i = %d\n",i);pq.push(i); // ans[++top] ='a'+i;}}while(!pq.empty() ) {int cur = pq.top(); // printf("cur = %d\n",cur);pq.pop();ans[++top] = cur+'a';for(int i = head[cur]; i!=-1; i=e[i].ne) {//是ne啊！！！！ in[e[i].to]--; if(in[e[i].to] == 0 ) {pq.push(e[i].to); // ans[++top] = 'a' + e[i].to; }}}if(top != 26) return false;else return true; } int main() {memset(head,-1,sizeof(head));memset(vis,0,sizeof(vis));memset(in,0,sizeof(in));memset(maze,0,sizeof(maze) ) ;cnt = 0;cin>>n;//從str[1]開始讀入字符串 for(int i = 1; i<=n; i++) {scanf("%s",str[i]);len[i] = strlen(str[i]); // vis[str[i][0] - 'a'] = 1;}int flag = 0;//點(diǎn) 從0到25； for(int i = 1; i<n; i++) {//是小于！ flag = 0;for(int j = 0; j<min(len[i],len[i+1]); j++) {if(str[i][j] == str[i+1][j]) continue;flag = 1;if(maze[str[i][j]-'a'][str[i+1][j]-'a'] == 1) break;//寫成continue了。。 add(str[i][j]-'a',str[i+1][j]-'a',0);maze[str[i][j]-'a'][str[i+1][j]-'a'] = 1;in[str[i+1][j]-'a'] ++;break;}if(flag == 0 && len[i] > len[i+1]) {printf("Impossible\n");return 0;} }if(topu()) {for(int i = 1; i<=26; i++) {printf("%c",ans[i]);}}else printf("Impossible\n");return 0 ;}

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